Answer
(a) $g'(0)$ does not exist.
(b) $g'(a) = \frac{2}{3a^{1/3}}$
(c) $g(x)$ has a vertical tangent line at $(0,0)$
(d) On the graph of $y=x^{2/3}$, we can see that the graph comes to a sharp point at the origin. As $x$ approaches $0$ from the negative side, the slope of the graph becomes more negative. As $x$ approaches $0$ from the positive side, the slope of the graph becomes more positive.
Work Step by Step
(a) $g(x) = x^{2/3}$
We can find $g'(a)$:
$g'(a) = \lim\limits_{x \to a}\frac{x^{2/3}-a^{2/3}}{x-a}$
$g'(a) = \lim\limits_{x \to a}\frac{(x^{1/3}-a^{1/3})(x^{1/3}+a^{1/3})}{(x^{1/3}-a^{1/3})(x^{2/3}+x^{1/3}a^{1/3}+a^{2/3})}$
$g'(a) = \lim\limits_{x \to a}\frac{x^{1/3}+a^{1/3}}{x^{2/3}+x^{1/3}a^{1/3}+a^{2/3}}$
$g'(a) = \frac{a^{1/3}+a^{1/3}}{a^{2/3}+a^{1/3}a^{1/3}+a^{2/3}}$
$g'(a) = \frac{2a^{1/3}}{3a^{2/3}}$
$g'(a) = \frac{2}{3a^{1/3}}$
Since $g'(0)$ is undefined, $g'(0)$ does not exist.
(b) As found in part (a), $g'(a) = \frac{2}{3a^{1/3}}$
(c) $\lim\limits_{x \to 0^+}g'(x) = \lim\limits_{x \to 0^+}\frac{2}{3x^{1/3}} = \infty$
$\lim\limits_{x \to 0^-}g'(x) = \lim\limits_{x \to 0^-}\frac{2}{3x^{1/3}} = -\infty$
Therefore, $g(x)$ has a vertical tangent line at $(0,0)$
(d) On the graph of $y=x^{2/3}$, we can see that the graph comes to a sharp point at the origin. As $x$ approaches $0$ from the negative side, the slope of the graph becomes more negative. As $x$ approaches $0$ from the positive side, the slope of the graph becomes more positive.
